 So we compute the various expectation values of each of the bi, gx, and the later things in the last class. We now want to use that to understand what the strings get. So how are we going to do that? Let's start with a simple example. Well, first I'm going to remind you of the formula we used in the game. It's very easy to find the i of yi, probability of the i, expectation value is equivalent to the world. yi in j, probability i in s in j, 2 alpha prime ai. And here is the right last statement. Okay, now let's study the three-tacton sketch. The three-tacton sketch is going to be a member of the rules, because we've got three c operators. So let's say a c operator and y1, a c operator and y2, and a c operator and y3. Times e to the power i k1 and y1, e to the power x at y1, e to the power i k2, x y2, and e to the power i k3.x at y2. Okay, now we want to understand the six-tacton sketch values. Well, the probability iteration is up, because the probability iteration is up. On season, the probability iteration is up. We can calculate the expectation value separately. And that's what we've already seen. It's y1, 2, y2, 3, y1, 3, we've computed. So that's the part we see. Now let's compute the probability axis. Okay? So the probability axis goes to this office. We've got e to the power, so we get y1, 2 to the power, y2, 3 into k1, k2. y2, 3 into 2 to the power i k2, k3, and y1, 3 into 2 to the power i k1. Now, from general principles, all three vertex operators are fixed at particular points, y1, y2, y3. Right? If we could choose these points to be any way, well, in this simple calculation, I'm not sure if they're in the procedure. Yeah, that's true. I'm not sure if they're in the procedure. I mean, it's consistent to check on our procedure that this the satisfaction should not end with the insertion point from the device. So first, it sounds crazy. I think it works. This depends on the momentum. Then you see the probability of this in a two-fold. And how? Remember, the k1.k2 is equal to the k1 plus k1. k1 plus k2 minus k1 plus k3 squared. k1 plus k2 plus k3 squared is equal to k3 squared. So each of these is given by the master value of the master value. So the master value here is k squared. The master value is minus 1 by 2. k squared is 1 by 2. So this is equal to minus 1 by 2. So yeah, 1 by 2, 1 by 2, 1 by 2, 1 by 2. OK? Sorry, I just had a 2. The problem was connected to that, with a 2. So we could choose the y1 to be raised to the power minus 1. Certainly y1 to be raised to the power minus 1, y2 to be raised to the power minus 1. And we can get the solution. There is a big truth, or a truth here, that in this case, the example is more less true than the amount of the additional factor of 2. But this is sort of the important thing, more degrading than me. And that's it. Now there's a moment in k1, k2, and k3. We place them at some point, y1, y2, y3. But it doesn't matter what this point y1, y2, y3 are, because we can choose them to be anything, by the appropriate, more based master, the appropriate, unfixed, conformable transformation that needs a lot of reaction. But, you see, any value of the change that acts in a one-to-one fashion on the power machine cannot change the relative ordering of points. Particularly on ordering. Now the way that we need to do the activation is to regard the, the version to the open string is the upper half place. So the upper half place can formally be related, it can formally transform over into the disk. And you can identify points at infinity, so they become boundary. So, the points that we put, such as y1, y2, and y3, if they're going to have a linear order, these points are going to cycle. So it's just cycling the ordering of y1, y2, y3, on the boundary of the world machine, that cannot be changed by quantity. So, if we want to sum over all the digital ways of inserting our graphics operators, we should sum over all the scientific orders. And if we're a bigger scientific order, we can choose this y1, y3, y3, and y3. Is this correct? Is this correct? Coordinate transformations and unfixed characteristics procedure that allow us to take y1, y2, y3, anything, have to be one-to-one. So, y1 cannot change the relative ordering of points on the boundary. Is this correct? Yes, that has to be summed over xx. Now, since this invisibility is a simple case, we have to do this, we do not care. It's not a question of y1, y2, y3, anything. So, if y1 were to be done in the past too, we're not doing this calculation that the length of the truth track and such. That doesn't drop in the end of the question. So, for this particular calculation, we would not care very much. That would soon become important in the end of the question. And it will also become important in this calculation when we generalize the calculation in a particular way. So, that's what we need to do in the end of the question. So, that generalization was a problem. So far, we've been talking about how open-stream is living on the strip. But it's not an amount of information that goes into the stream. Once you have boundary conditions, you've got some special boundary points. You might, you know, where is the consideration to be dealt with? That's not the propagation of the theory it has been in the past. That we can't change easily since it's very, very time-consuming. However, you might, once you add, you boundary use a tree. You increase the freedom that lives just at the boundary of a tree. The question of how you can add new boundary degrees of freedom in a way that preserves the formula in the end and all the other good properties of the problem. I think that is an intricate and interesting question. Well, we will not try to give a harder answer than that at all. All you've got to do is to make very... No, you'll do better. All you've got to do is to make very complicated charts. Okay? There are other ways to do it and that makes it more complicated because I've got boundary states, so they're there already. At some point, of course, the research will end up with this. But as a matter of fact, you've got the very trivial set of boundary degrees. The boundary degrees of freedom will only have to be given a vector space that is the final dimension. Okay? So there's a final dimension vector space during each boundary. Okay? And we work in some basis to make this space. So there's some inner problem of the vector space, the inner space, and we work in some basis, some common basis, we use the basis vectors to recall them. Okay? So the basis vectors are, you know, column one, zero, zero, zero, zero, zero, one, zero, zero, zero, and so on. And then the only difference is boundary degrees of freedom. The evolution of the evolution of the boundary degrees of freedom will be true. Okay? Which means that along any given boundary the boundary degrees of freedom don't change. Okay? Basically because, basically what I'm saying is that it's that the open strings in the picture come in different flavors. But flavors are labelled by an index, one index i, around from one to n, where there is anything. One index i, around from one to n, on the left boundary, around from one to n, on the right boundary. Our open strings on the strip, in addition to being labelled by, you know, the full Hilbert space that we obtain by quantizing these open strings is specified by a vector and the tensor product. After Hilbert space we have before, I mean the oscillator is in zero mode, and this n-square dimension Hilbert space it can have some boundary degrees freedom on the left. Okay? So you can think of it as some matrix, n for the matrix, times wherever we have the form. Okay? Think of it as n-square different species of open strings. So n-square different species of open strings. Okay? The species does not change in evolution. Right? They can change in interactions. Each species, because identically, I don't know how to think. Is this clear? I cannot see very systematically what you mean today. There is a very consistent generalization. It's one of a much larger, there's a full theory one can build up, consistent boundary states of boundary interactions one can put to a straight. I won't stop to try to build this volume. We just look at this as one example. It's something that's more or less trivial and therefore when you look at it carefully more or less clearly consistent with the generalization of the volume of this normal n-square condition. Okay? So now the important thing is that states of the open strings and then vertex operators are labeled not just by the vertex operator label links that we've got before. These are by IPX and derivatives and so on, but are also by n-square different species. That can also be a wide line, a long, long one, just not n-square different species. So the index cannot change at all. So if you have an open string and its index cannot change at all, however it comes and interacts with two output strings. So the index around here is the insert. The index around here is the insert. The index around here is the insert. Okay? Now these open strings are replaced by the vertex operator. Any computation involves the insertion to your whole vertex operator. It tells us that you have to put in addition to what you had before, you also have to put in the trace of yawnawa yawnawa yawnawa yawnawa yawnawa. No, it is actually the three cations cations that we have with you. The three cations that we have with you. Anything in the large sphere depends on these new boundary decreases. Okay? So let's do that. All we've already seen would be that the answer is one from the matter part. Okay? So first we had lambda 1, lambda 2, and lambda 3. So this would be the pi k x1 in the pi k x2 and in the pi k. k1, k2 k1, x2, x3. These are our three values. We've already seen the expectation of what orderly we choose. Okay? We get one from the matter part and matter goes up to the end. So let's do the precisely what we've already seen. So let's put in the random three or one, three, and two. So these two they should try to get all the numbers. They don't give us equal boundaries for the lambdas. Because one of them makes us place of lambda 1, lambda 2, lambda 3. And the second one we first place of lambda 1 lambda 3. So we want to do, we can write this as for instance, place of the anticognitive of lambda 1, lambda 2, lambda 3. Times 1 belongs to the three tachyons that we have to use. Okay? Where the tachyons are labeled where n cross i means matrix, but the x operator lost a point. And the answer to that is we're going to do another three-point function calculation. This time of this calculation two tachyons will be what is the physical interpretation of this calculation? If I take two tachyons of the tachyon what will help me learn whether the tachyon is charged or how it's charged under this case? Well, if we take the tachyons of the tachyons of the tachyons it's the natural gas of this gas. And the gas goes on to be non-immediate. Okay? And the tachyons will be non-immediate tachyons. Okay? But let's start seeing. So let's start seeing. So I'm going to do this to compute. Okay? So the three operators of gas, that's how the gas goes on to operate at number one. So we have e to pi k1 dot x. And then it's associated with the polymerization factor e dot x. And let's start down with this matrix signal level. Then we have e to pi k2 dot x. This is associated with the tachyons. And e to pi is e to pi k3 dot x. Again, this is a root. e to e to pi k3 dot x to tachyons, that's how it works. Or the matrix. It makes a problem. That there is a problem. Also the C3 matrix. It can be done once again. We can fix that. Okay? Now let's compute the C3 matrix. So once again we have two orders. We have the ordering let's say one, two, three. And the ordering one, three. Again, let's do the calculation first for the first. Let's do the calculation for one, two, three. Okay? First thing for the C3 that's the first one, y1, 2, y1, 3, y2, 3. Okay? And that will be common for both buildings. We'll add that on the side. Okay? Now there's also the calculation for the loudness. So the first guy gets I'll try to do the calculation more or less. The first guy will get a trace of lambda, lambda. Well, we have to distinguish between the traces. Let's say we take up traces of that. I'm not sure if you should dimension this but let's say that we would take up traces of this. That matrix is one. Okay? So, lambda1, lambda2, lambda3. And this guy of course, and that would be something other than it. This guy would have a trace of lambda1, lambda2. So we have to figure out what model class is it and what model class is it So that's what we have to do. Now we get by contacting this guy and this guy and this guy. In order to do that we need the rule that we developed last class. The rule that we developed last class helped us deal with insertions of this data. So the rule that contracts the exponential size of the data for the data and adds the additional factor of the data. Okay? So of course let's do the contractions of the data and then you are going to be using that. So let's look at the number that we got, that's y1. So we have y1 and 2 to the power over. Well, it's 2r times k1 dot k2. Now once again, k1 dot k2, 2k1 dot k2 is k3 squared minus k1 squared minus k2. k1 squared is 0 and k3 squared is equal to k2. So we get 0. 2k1 dot k2 which is k3 squared minus k1 squared minus k2 squared k1 squared is 0, k2 squared is equal to 0. And similarly we get y1 3 to the power of y2 3. For y2 3 we get 2k1 squared so we get k1 squared minus k2 squared minus k3 squared. k1 squared is 0. But k2 squared is k3 squared at each one by alpha k. So this is minus 2r times k1 dot k2 but we get the formula for 2k1 dot k2. Any reason that earlier it was minus 1 this is now. Now what do we get from the insertions? So what do we get from the insertions? So from the insertion we get if you remember from the formula it says we get 1 over there is a number where 1 over y1 minus y2 into k2 plus 1 over y1 minus y3 into k3. The general formula was check where this guy is inserted this guy is inserted into the formula. Just take yi minus yj to all other insertions and have the k of the other guy. The guy is going to be dotted with the polarization tensor. Yes? Okay. In this case, can we make an insertion? Nothing more. Now let's process the formula. This formula idea has what? We have k2 into k3. The first thing we have to find out is that k3 k1 dot epsilon is 0. k1 dot epsilon is 0. That is one of the physical state conditions the BRSP closed conditions of our experiment. Therefore by the way the conservation k2 dot epsilon is minus of k3 dot k1 plus k2 plus k3 is 0. Each of these, this k2 will be replaced by k2 minus k3 by 2. k3 can be replaced by k3 minus k2 by 2. So this is equal to epsilon dot k2 3 by 2 into 1 by y1 2 1 by y1 which is epsilon dot k2 3 by 2 into now we get y y2 3 over y1 2 y1 it's y2 minus y3 y1 2 y1 less. So now let's all work together. The first thing we have to take into consideration is that everything is going to be in the length of the each section. That part is going to work. That part is going to work because that part is going to work because now we got this times y 2 3 to 1 by 1 by y2 3 2 k2 1 by y1 3 y1 3 y2 3 and that cancels these sections of the series. What we get from the other one but notice that the other order is just obtained by replacing 3 by 2. Replacing 3 by 2 is epsilon dot k2 3 into lambda 1 into lambda 2 up to epsilon. I would like each of you to go home and make this minus sign to epsilon dot k. The method under these two order is to contribute to the relative minus k. That's the thing that's very important. Now to reach this commutator that has to be no polarization tens. The lambda has to be trivial. Where is the minus sign? k3 2 is minus 2. If lambda has to be trivial this would have just been 0. So we would have in the statement that the tacking on is uncharged under the gauge force. In the general case it's not trivial. But we also see how this thing this is the the basic structure of the two categories. It's the commutation that appears. The commutation that appears multiplied by the matrix of the gauge force. Now this is in fact the statement that the tacking on is interactive with the gauge force are exactly like a minimally coupled and joint scalar interaction matrix. What is that? That's why there's a Lagrangian minimally coupled action. The Lagrangian minimally coupled action is a stress of dv5. That's dv5. Two tacking of the magnitude that goes on interacting from comes from one of these d-news having the area. And the level of that is commutated by dv5 twice. And this one has the odd derivative. This is exactly the same as the swamp structure. You know where we put the commutator doesn't matter. It would be uncoordinated by many tools. It's the same, let's say. Where we call it one and two words. Just after we put two cyclical odd bits with a relative amount of time. And then this is exactly the scalar structure. The derivative here is the fact that the has a k. And the fact that this is a derivative contract to be a u is the fact that the k contracts to the polarity. You know, we've got a little more detail than we have here. It says that the sign of the rule of the scattering amplitude that follows from this is the factor. But the open string theory I mean, it's really seem very much like open string theory has not a billion gauge bosons there. And the datum's is charged like an adjoint scalar under least now. Look at our specific theory of datum. Okay? Every field in the problem are the gauge bosons setters. But every field in the problem enters as an cross-end. So it's plausible that every field in the problem is charged in the adjoint of this, of this gauge. So what I'm saying is that if you work out the first state sign of the state, two copies of that scattering against the datum once again you find the derivative. If you're detecting that basically all states of the theory are charged in the adjoint of this u and d2. Okay? The one reason that expression in this regard is the gauge boson itself. Expression because it should give you all the interactions and if our derivation is correct, it should give you the scattering amplitudes that follow a problem. That follow a problem the interactions of gauge bosons itself in the y-axis. So something that's interesting to do is to calculate the 3-point the 3-point spectrum of 3 gauge bosons. Okay? And then later that makes a sense. It's more of a system. Yeah, you just have to do all this stuff again and again. If one new element is set as a rule for a contraction of this stuff, the other element is you remember there were also the oscillators, the active oscillators in one contract. You just have one derivative and there's nothing going to contract. Yes, we didn't have to use that. So now we have three derivatives. So there will be an oscillation of contractions fair-wise in addition to taking the derivative that is just this piece for everything that exists. Okay? And then there will be one term when you take this piece for everything that exists which will be Q e to 1. And then there will be three other terms which will be the area. Where you take this term for one of the two delixes and the oscillators will be out. Is that clear? Do you know what I'm talking about? Do you know what I mean when I say the oscillators? Do you remember what our rule was when we contracted these delixes? There's a part that contracts that's what's going to be our oscillators. Okay, the contract therefore only adds other elements. And there's a part that has background value that will be the Q impact. The background value is a linear domain. So the term when you take all the three background values will be Q e to 1. And the term when you take one parameter, the group delixes come back to Q e to 1. There will be four terms because if the oscillators can choose there are three ways to choose it in which space they come back to Q. Okay, so there will be four terms that have to be back up. Let us go to the result of the purchase. So now together apart from these very special functions on Q e to 1. Q e to 2. Q e to 3. Plus sign. And then there's a special job which is just I'll go back to Q e to 1. Q e to 3. Q e to 2. Q e to 1. Q e to 3. The rest of lambda 1. Now as interval. The doctor tells us that interval no matrices because it's omni-ray. We were just in U1 carry. We get 0. Which is great because gate was also interacting with itself. You interact with each other. I'll simply define the rules that come out of the usual question. The three-point line of notes are done. You should check it. It's different for instance starting from the fact that it is three-dimensional in one day. And it's different also that it has an extra alpha minus. And it reflects the fact that string theory is language. String theory. For the masses, increase or decrease. Reduce the degree of radio-energy. But as an extra interaction that are not there in the middle of the problem. It's correct. The rest of this is that to introduce some I don't have a ground signal. It's proportional to alpha prime into 2 mu f mu alpha f alpha. For two of these f's and the linear is from alpha, yes? Yes, just the linear is from alpha, yes? And that is the rest of this. Why is that? Why is that? Let's see. So, this is effectively L mu A mu minus L mu and similarly L mu alpha L mu A alpha minus L mu. And why do we get this structure of interaction? Let's see. I have to say I have to check this. I'm just going to use the question scheme. As for prime, this is very much important. No, the other prime is that that could occur. And that is the other prime. We see that. We see that and you say that this thing would just be non-zero. Yeah, you're probably right. I'm sorry, I have to check this. You're probably right. Yeah, it's probably right. Okay, let me use this in exercise. Let me produce this thing. It's going to have a 3D edge. It's going to have this contract. You're almost 79. How are we going to get this structure? How are we going to get this structure? Let's see. No. You see, I don't know this. You see, you see, this is supposed to be incompatible with ui, that's stressful. We got q-ray, we got an anti-selective. And that's probably just we got q-ray. Yeah, so maybe this thing doesn't come with ui. No, no, no, okay. I think that's what I think it's going to be the same with ui. Because of that. Exactly. Because the symmetry part is symmetry in ui. Exactly. So even if you're not there, this would just happen. So it's consistent with this kind of thing. This thing will be reduced by the term of being the such a term effect in the ground. I have a check, but it's something like this, but I do the job, check it out. Please do the exact same thing. Okay? And try to convince yourself that what you get from here is exactly what you expect from the ordinary industry. And what you get from here is that the term is not inconsistent with string theory. It could be a gauge theory of learning. Because if all the derivatives are large in the ground, this term is much smaller compared to this. This has two issues there. One is the power of alpha, compared to this term. So even if the derivative is large in the limit where the length is greater than large in the U.S. ground, they are there, but they matter only for configurations of varying of length scales or of energy scales or all of the string scales. If we have a varying of length scales large compared to string scales and it's lower compared to string scales, that term is negligible compared to string scales. It's the scale. No energy to get back to string scales. There are connections around it. It's very even string scales. But we know it's very simple. It's not the only girl. This is the effect of action that comes from some. You know, some of it over all our string theory. This is one down in the effect of action. The first to effect of action. And we have independent ways of understanding that the string theory is very finite. So those issues aren't on the issue. It's just basically the same. When you're starting to encroach in theory, you have the gravity. And then we have high level of interest than the gravity but the whole theory is finite. The regularization that makes sense. We should expand it out in the lattice space and increase it. Increase it higher and it helps. The whole theory is nice in theory. And the manager can start a very good one in the string theory in theory. It should do something like e to the power minus there is squared and there is 4. That's the fact that the suppressor is high. It will go worse and worse as you're going to higher levels. I know you're very normal all the time. Yes, it's certainly not easy to make consistent theories of gravity, especially gravity by hand, doing something. So string theory manages it in a very beautiful and intricate way but I do not include it like that. But just if you want a rough analysis of it, string theory isn't all about gravity. So, I mean, it's not not local. But at least there's no string theory. There's no such thing. What do you mean by the gravity in theory? I don't know anything. It's to move on to the discussion of four points. So we're going to just move on to the discussion of the let's see how it will happen. It's actually the first formula of a string theory that is string theory. And you get to understand this beautiful formula that likes to let it pass to the world. Okay. So, it's not normal that I mean, there should be the other type of string theory that's possible in one sense when we work on it. Even if one of us has one. But there's really direction to all of us. Yeah, but they don't appear to have three points. They don't appear to have three points. Three points. Yeah. We're going to have to get the three points. We've always got to get the three points. It's exactly the two points. There's no further reason. Of course, even the three points would get the electric quantum mechanics. What's the difference? Is this the most general string I know? No, this isn't the most general string but, oh, you're asking this. Just from a word, okay, let me get this one. Yeah, I see that any little more, more Fs would necessarily come from your only hand. Yeah. They have four Fs. They have four Fs. Yeah, so maybe you get this one. Basically. No, I mean, you could get me that string theory. Yeah. I mean, you could get some square root three times our problem. I mean, it's very... Okay, now, in this question when we do this, the same thing in so many terms that could appear wrong. Okay? But we understand a lot of that from the first time we've seen this. From the first time. Okay? But certainly, if you were just looking at things without any differences, that could appear wrong. Yeah, but there's obviously any string there. If you have a particular answer, although if it's a particularly particular, it's something particular. So, that's a lot of the aspects. Let me ask. Okay, and this is what I think is, whatever it is, the question is, I mean, to leave relations between this and other things. Yes, yes. That's what we do. Right? And that's the next time we're going to do it. Now, what's the location of that coming up? But let's just this one, you know, let's just see the interaction. Okay, now, let's just quickly move on to the four points. The four points that appear wrong. Now, as you remember from our first question, it is usually possible that in a delegation that is not just writing a number but in case. Right? Because on the version, you mean? Because you can fix three of the four vertices of the fourth one, is where it reaches. Right? Let's set up the aggregation to that. So, we've got let's say four vertices of the fifth in one at x, it's a lambda one. It's pi k two dot x, it's number two. That's all. I'm thinking about it that for formal reasoning, we can set those three points to be wherever we want. Let's set them to be the main interests. Let's set them to y one equal to zero, y two equal to one, y three equal to zero. So, this y i k is from C's we have. y one two, y one three, y one four, y two three, y two four, this is the number of C's. Okay. Now, anything that does not involve y four is just a number. So, the only, so the terms that involve y four are y one four, y two four, y three four. And then there are some numbers. The numbers are like y one minus y two is one. But anything equal to y three is infinity. We have infinity to the power. Okay. So, that's this is also infinity. It's infinity to the power. You know, more precisely, it's y three to the power. Three is the limit of y three. Let's see what we get from the momentum market. From the momentum contractions we get, we have e to the power k one three eight and so on. We got this product of i, right. We got e to the power two alpha prime, k i dot k g. From the point. Let's see what k i dot k g is. What this k i dot k g is is this. Sorry, y and j. Let's just add those terms that involve y three. All such terms are infinity to the power two alpha prime into k three dot k one plus k two plus k four. And we say this as well as y three is basically y three to the power. So it's infinity to the power of this. So that's the infinity to the power minus k three squared. Minus k three squared was minus one alpha, right. So that's infinity to the power minus c. That's what I'm talking about, right. With the c's, only three are fixed. Only those three c's are here. Why won't you y two three, y y one two, y one three, y two three. That's how it was formed. Here we didn't get it, guys. And what we did get was infinity. So if you take it over, that doesn't have c's. So I'm talking about that. So there's infinity to the power minus k three squared. So what's left is all the stuff that involves y four. So what do we get? The stuff that involves y four is okay, once again there's a y one two. That'd be okay. So the y four into k one dot k k four and y four minus one into k one dot k two dot k four. Okay. Now, there could be some time. Due to that decision, s is equal to k one plus k two, that would be squared minus time. And there could be four components. It's kind of non-dimensional. Alpha prime t is equal to minus k one plus k three. It's where alpha prime u is equal to minus k one plus k four. And this object, k one dot k four is equal to so two k one dot k four with alpha prime is equal to alpha prime alpha prime k one plus k four double k squared minus k one squared minus k one squared alpha prime is equal to minus u minus k. So that object there is so our answer involves the d y four y four to the power minus u minus two y four minus one to the power and then this would be similar except we have x minus x. So that's the kind of detail that we have to do. That's the kind of detail that we have to do. If you want to solve this problem. Now, one last thing that we stopped at was this y four is integrated over what? It's integrated or minus integrated. But you see, what's really important to stop. We have y one, y two, y three except that zero one is this object and the integral here from in this way from zero to one which was the same answer as the integral here to one thing to delete just because we couldn't prioritize it. The only way to integrate is what? Order the various vertex operators inside. So instead of returning this integral as going from minus infinity to infinity we can think of three different integrals. One from each range minus infinity zero to one and one to infinity and the part that is from minus infinity to zero which has the same answer as one to zero to one so why do we integrate vertex operators in a product? Actually there are six different combinations. Each of them is proportionate or very similar in terms of integrated vertex operators. So the six different combinations are the following. Y four here or y one, y two, y three and y four here or y one, y two, y three and y four here. Y one, y two, y three because of the thickness I could do it. We could have had y one, y three, y two and y four y one, y three, y four and y two and y one, y two, y three and y four. y four go in this range in each of these paragraphs and each of them will be less than one to zero to one where the appropriate vertex operators desert desert. So what I'm making and I'm going to make that quite probably just to give you the answer is that the answer is some of six different terms. Each of which is an integral of this form from zero to one. In different terms this power is used in excess of energy. In different terms and k one dot, we have k one dot k four here and k two dot k four in these two rectangles because we were going, y four was going between one and two. So k one dot k four and k two dot k four when y one goes between two and three this would be replaced by k two dot k four and k three dot. What each of these is done is used at six to seven rectangles each of which is the same form compared to x k. Now the other thing to remember is that by the variable change y four goes to y four minus one for one minus one four we should have discriminated. So this is symmetric in the interchange of x and u on the interchange of one and two. So that's not what order this through. It's a choice of u and x. So the three times are not the invariance x, t and u okay and we get six terms two, two of which will have s and u, two of which will have x and t, two of which will have t. And what we have to do is the thing is we have to keep track of which diagram gives which dependence so that we keep track of the lambda. So that's two is one. Two is one. Let's find out. This diagram you already see four which is u and one and two which is four and two four and two which is the same as one and three. Yes, why do I have to do this? This was one three. Yeah, this was one three. You and me. What about this side? This is now going to give us two and four which is the same as one and three. And two and four and three which is the same as one s. This is going to give us four and one so that's u and four and one so that's u and four and three that's s. So that's one and two. This has y four, y one by three which is u and s. This has y four by two the same as one and three that's t and y four, y three is s and I'm here so y four, y three which is s and y four, y one which is u. I guess we've got two s's now let's do this again. y four and y one which is u y four and y three which is s this one is y four and y three is s and y four and y one which is u and this one is this is two and three. Three and two. So this one is y one, y four which is u and y four and y two which is t. We've got two s's two s's so that thing one and four is a u one and four is a u yes this one is u one and four is a s one and four is a u u t okay, we have two ut's ut ut we have two s's s t s t and two u s's okay, that's the problem okay, so now, suppose we write down this the answer, whatever this is as e of s t s t s this thing is multiplied by what? trace of, how do you look at those that give us s t s t s t lambda one, lambda two lambda four, lambda three plus lambda one, lambda four that one is lambda three lambda four lambda two. So the two diagrams give us s e of ut and then the second order is two diagrams and give us u t so u t at least two we get one, four, two, three plus one, three, two, four a of s u which is one, two, three, four plus s of one, four, three okay, we continue next time because we have to derive the differences actually, for t t, that's wrong okay, we continue next time it's amazing if we want to know how we